Chapter 15 Acid-Base Equilibria

Solutions of Acids or Bases Containing a Common Ion—Section 15.1 A buffered solution is made by combining a weak acid and it’s conjugate base Buffers take advantage of the common ion present in solution to “buffer” against any changes pH Ex: Acetic acid and sodium acetate solution

Calculating the pH of a Buffer What is the pH of an acetic acid/acetate buffer made with 0.4 M sodium acetate and 0.20 M acetic acid? Ka for acetic acid is 1.8  10-5

Henderson-Hasselbalch Equation for Buffers Calculate the pH of the previous acetic acid/acetate buffer using the equation above Can only be used “when x is really small”

Preparing Buffer Solutions Calculate the mass of ammonium chloride that must be added to 500.0 mL of 0.32 M NH3 to prepare a pH 8.50 buffer, Kb for NH3 is 1.8  10-5.

How Buffers Act: The Addition of Strong Acid or Strong Base The buffer capacity for a particular buffer refers to the amount of acid or base that a buffer can neutralize before large changes in pH occur Ex: 1 M HF/F- vs. 0.1 M HF/F- Both solutions have the same ratio of acid/base; however 1 M solution has the higher buffer capacity because it is capable of neutralizing more acid or base

Acid-Base Titrations The process of systematically adding acid or base for neutralization is known as a titration Note: Titration is a technique and can be applied to other types of chemical reactions besides neutralization reactions

Titration Curves

Strong Acid/Strong Base Titrations Calculate the pH in the titration of 20.00 mL of 0.125 M HCl after the addition of (a.) 0 (b.) 2.00, (c.) 10.00, and (d.) 20.00 mL of 0.250 M NaOH

Titration of a Strong Acid w/ Strong Base Calculate the pH at each of the steps indicated below for the titration of 200 mL of 0.15 M HBr. After the addition of 0 mL of 0.15 M KOH After the addition of 100 mL of 0.15 M KOH After the addition of 200 mL of 0.15 M KOH After the addition of 250 mL of 0.15 M KOH

Titration Curves for Strong Acid/Strong Base Titrations Divided into different regions Region 1: No base (pH << 7) Region 2: Addition of base up to equivalence point (pH < 7) Region 3: Equivalence point; indicated by the inflection point of the curve (pH = pKa) For strong acid/strong base titration, pH = 7.0 Region 4: Increased addition of base after equivalence point (pH > 7)

Titration Curves for Weak Acid/Strong Base Titrations Also has 4 regions although the regions are slightly different Region 1: Presence of weak acid only Not 100% dissociation so ICE tables must be used to determine pH Region 2: Buffer region Weak acid + conjugate base (salt) = buffer; use Henderson-Hasselbalch to calculate pH Region 3: Equivalence point Equal amounts (# of moles) of acid and strong base; conjugate base is produced and Kb must be used to calculate pH Region 4: Addition of excess strong base Because excess OH- is present, it will dominate the pH; amount of conjugate base is irrelevant [OH-] is used to calculate pH

Titration of Weak Acid With Strong Base Calculate the pH during the titration of 20.0 mL of 0.500 M formic acid (Ka = 1.8 x 10-4) with 0.500 M NaOH. Calculate the pH at the following intervals: 0 mL NaOH 10.0 mL NaOH 20.0 mL NaOH 30.0 mL NaOH

Titration Curves Strong Acid w/ Strong Base Strong Base w/ Strong Acid

Titration Curves (cont.)

Polyprotic Acid Titration Curves Diprotic Acid Triprotic Acid

Section 15.5—Acid-Base Indicators Acid-base titrations are best observed using a pH meter which will give a digital readout of pH throughout the course of a titration Acid-base indicators can be used however to at least determine when a titration has reached the end point

Solubility Equilibria In reality, there is no such thing as a 100% insoluble compound Only practically so Example: AgCl in H2O Insoluble compounds undergo an equilibrium in the same manner as gases, weak acids, or weak bases Equilibria constants for molar solubility are represented as Ksp

Calculating Ksp When lead iodate, Pb(IO3)2, is added to water, a small amount dissolves. If measurements at 25 C show that the Pb2+ concentration is 4.5 x 10-5 M, calculate the value of Ksp for Pb(IO3)2. See Interactive Example 16.1 (Pg. 642)

Calculating Solubility from Given Ksp Values Given the value of Ksp for BaCO3 (Ksp = 1.6 x 10-9), calculate the solubility of Ba(IO3)2. See Interactive Example 16.3 (Pg. 644)

Solubility and the Common Ion Effect Assuming the following equilibrium: CaF2(s) Ca2+(aq) + 2 F-(aq)

Solubility and pH Essentially just a special case of the common ion effect Ex: Solubility of Mg(OH)2 in neutral vs. basic solution

Calculating Solubility in the Presence of a Common Ion What is the solubility of calcium hydroxide (Ksp = 5.0 x 10-6) in 0.050 M NaOH solution? See Interactive Example 16.4 (Pg. 648)

Predicting the Precipitation of Ions A chemist mixes 200 mL of 0.010 M Pb(NO3)2 with 100 mL of 0.0050 M NaCl. Will lead chloride precipitate? Ksp = 1.7 x 10-5 See Interactive Example 16.5 (Pg. 650)